YEAST GENETICS: FROM GENES TO PROTEINS TO BEHAVIOR

Table of ContentsFlowchart of the FGPB project-------------------------------------------------------------------------------- 2Objectives of the FGPB project------------------------------------------------------------------------------- 3Adenine synthesis pathway (specific names)----------------------------------------------------- 4Purine metabolism pathway--------------------------------------------------------------------------------------- 5Can ultraviolet light change the appearance of yeast?------------------------------------- 6Characteristics of red yeast: Can they be obtained by treatment of whiteyeast with UV light? Do they have special nutritional requirements? ----- 13Preparation of DNA from a red yeast strain-------------------------------------------------------- 19DNA from the red yeast strain: Electrophoresis and PCR---------------------------- 23Analysis of the base sequence in the DNA of the ADE1 and ADE2 genes

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FLOWCHART OF THE FROM GENES TO PROTEINS TO BEHAVIOR PROJECT

Laboratory Session

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Activities Determine iftreating a whiteversion of yeastwith a DNAdamaging agentwill give rise tovariant forms(e.g., a redversion).

Observecolonies.Test thenutritionalrequirements ofthe red type andthe white type ofyeast.

PurifyDNA from the redversion of yeast.

Amplify ADE1and ADE2 genesto allow fordetermination ofthe sequence ofbases that makeup these genes inthe red version.

Use an electronicdatabase toidentify featuresof the ADE1 andADE2 genes ofthe white versionof DNA and compare to the genes from the red yeast.

Learning Objectives for the From Genes to Proteins to Behavior Project

During this unit, you will perform a series of related experiences that mimics the process of investigational science and is aimed at accomplishing the following learning objectives.

(1) To understand the Central Dogma---how information "flows" from DNA to RNA to protein(2) To understand how the sequence of nucleotides in a gene provides the information to make aprotein(3) To understand how the sequence of amino acids of a protein determines the structure of a proteinand to recognize the relationship between the structure of a protein and its function(4) To understand mutation, its random nature, and the relationship between mutation and variation(5) To understand how a series of chemical reactions, each one catalyzed by a specific enzyme, can result in the production of essential "building blocks" of the cell and to understand theconsequences of an inability to carry out one of the chemical reactions in the series

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Laboratory 1: From Genes to Proteins to Behavior

CAN ULTRAVIOLET LIGHT CHANGE THE APPEARANCE OFYEAST?

In the From Gene to Protein to Behavior Project you will be studying baker’s yeast strains that differ in appearance. Yeast, schmeast, you may say, why should I care about different kinds of yeast?However, the presence of different types of yeast leads to a fundamental question of biology: What are the underlying causes of biological variation? Why are some yeast red and others white? Why do some of us have black hair and others blonde or red or brown? In the laboratory sessions of weeks 1-5 our overall goal will be to determine the cause of the variation we see in red and white yeast. The insights obtained by this study of yeast can be applied to understanding variation in other organisms, such as humans.

Specifically, in this first laboratory you will test whether one type of yeast can be changed into another by exposure to UV light. You will examine the effect of different doses of ultraviolet (UV) light on S. cerevisiae (yeast) cells and the resulting colonies. One macromolecule that is known to undergo change after UV light exposure is DNA. Can change in the DNA result in a change in the appearance of the resulting yeast colony? That is, starting with a yeast that forms white, smooth, round colonies and exposing it to UV light, will we see any colonies with altered color or morphology?

To most fully measure the effects of UV light you will subject the yeast to increasing doses of UV light. Not only can we answer the question of whether one type of yeast can be generated from another but we can answer more general questions as well. For example: What happens when macromolecules in a cell are subjected to chemical change beyond what normally occurs inside of the cell? Does the cell grow faster than in the absence of the chemical change? Does the cell die as the result of the chemical change? Does the cell acquire a different property as the result of the chemical change? Do all cells react the same way?

Some useful vocabulary:sterile: adj. Containing no living organisms.medium: n. What bacteria or yeast grow in or on. Can be solid or liquid.culture: n. Yeast (or bacteria or other microorganism) growing in or on a medium.colony: n. Usually found on a petri plate. A visible "pile" of yeast (or other microorganism) that are all descendants of a single yeast cell.strain: n. A particular variety of a microorganism. Different strains are more similar than are different species. This term is functionally equivalent with the term “breed” (used in animal husbandry) or “variety” (used in agriculture). In this lab you will be working with a white strain of yeast to see if you can alter it to make a different strain with a new appearance. Do not confuse this term with the word “strand” which is a long thin object such as a hair or one half of a DNA helix.aliquot: v. To subdivide a liquid sample into smaller portions. n. A portion of a largersample.

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Methods Sterile technique: Remember, you are trying to see if you can change one type of yeast into

another by using UV light. If you see any colonies that appear different from the colonies you started with following UV treatment, you want to be assured that these colonies were derived from the yeast in your original culture and are not some random cells from the air that just happened to land in the tubes or on your plates. In other words, it is essential that your yeast cultures be kept free from other microorganisms that are found in the air, on your hands, or elsewhere in environment. You will use sterile technique to keep the cultures from becoming contaminated. There are several different components to sterile technique; your instructor will demonstrate those that are necessary for today’s lab.

Spread plates: You will want to be able to see individual colonies of the yeast, in order to see if any of them look different from those of the original strain. A spread plate is often used to separate the cells in a liquid culture from each other by spreading them evenly over the surface of a petri plate, so that each cell can give rise to a separate colony. The instructor will demonstrate this technique and instructions are also provided below.

Procedure

1. Obtain four YPD plates. These YPD plates are a nutrient-rich medium for growing yeast.

2. Obtain a liquid culture of yeast cells marked “HAO”. HAO is a strain of white-colored yeast.Check the label carefully – it is critical to use the correct culture! Just before using the culture, vortex it to mix.

The yeast cells are dense and will settle to the bottom of the tube. Thus, to ensure pipetting a homogeneous solution of cells, vortex immediately prior to each use.

3. Take a 0.1 ml aliquot of the culture using a sterile pipet and sterile technique. Put it in the center of a YPD plate.

4. Use a sterile spreader to spread the drop of culture as evenly as you can over the entire surface of the petri plate. If you are using individually-wrapped plastic spreaders, you can use the same spreader for all four plates, then dispose of the spreader in the biohazardous waste container.

5. Exposure of the cells to ultraviolet light: You will expose each of the four plates to a different amount of ultraviolet light. UV light can be measured in a unit known as a microJoule. Expose one plate to 3000 microJoules, one to 6000 microJoules, one to 12000 microJoules and one to 24000 microJoules. Label the bottom of each of the four plates with a marker to indicate the amount of UV exposure the plate will receive and the lab group’s initials.

Exposure to the ultraviolet light will be done in an apparatus called a Stratalinker that looks like a microwave oven. This can be programmed to give a known dose of ultraviolet light.

6. The apparatus that delivers the UV light has an LED that can be set. The readout represents the number of microJoules X 100. Therefore if you want to deliver 3000 microJoules the display should read 30. Also make sure the apparatus is on the “Energy” mode and not the “Time” mode.

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7. After setting the apparatus, place the appropriate YPD plate in the apparatus and remove the lid. UV light will not penetrate the plastic lid!

8. Press the “Start” button. It should take between 5 and 30 seconds for the UV light to be delivered.

9. Dilute culture for untreated control plate: As a control for your experiment, you should make a plate of yeast cells that is not exposed to UV light. The problem with this idea is that it would be useful to be able to count the number of colonies on this plate next week, but the culture we gave you is concentrated enough that there will be too many colonies to count. Therefore, you will dilute the culture before using it in the next step to make your control plate. Dilute the cell culture as follows: Mix the culture thoroughly on a vortex mixer just before you begin the dilution process. Using sterile technique, add 1 ml of culture to a tube that has 9 milliliters of sterile water. Vortex to mix the dilution. Be sure to mix thoroughly. Don’t forget to use sterile technique. Don’t touch the pipet tip to anything that is not sterile, such as your fingers or the benchtop. Using sterile technique, immediately pipet 0.1 ml of this diluted culture onto your new YPD plate. Use the same technique as described above to spread the culture.

10. Incubate cultures at 30°C: Place each of your five YPD plates face down in the location indicated by your instructor. It will be convenient if you tape the bundle of plates together. The instructor will place the cultures at 30°C, an optimal temperature for the growth of this organism.

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Lab Assignment 1Due at the start of Laboratory 28 points

Name:

(2 pts) 1. The following observation is made in comparing a plate that was exposed to UV light to one that was not. Explain how you would experimentally determine if the colored colony resulted, not from UV treatment, but from being in a crowded condition (nearby so many colonies).

(6 pts) 2. A biologist wanted to understand the quantitative relationship between the amount of UV light (in microJoules) used to expose a culture of yeast cells and the percentage of yeast cells that survived the treatment (where 100% is the number of cells surviving after no UV treatment). She obtained the following result.

Amt of UV light (mJ) Number of cells surviving % of cells surviving0 1,540 100300 1310600 65900 5401200 91500 8

(a) Fill in the blank spaces in the table.(b) Use the computer program Microsoft Excel (or your favorite graphing software) to graph theabove results in a scatter graph. The plot should show the percentage of cells surviving as afunction of UV dose. Refer to the section of this manual named “Using Microsoft Excel toCreate Graphs” for general instructions. Label the X and Y axes and place a title at the top of the graph. The title should be meaningful but not verbose. For instance, a title of “Yeast experiment” will not have any meaning for anybody who is not in Biology 131 in the spring semester of 2004. On the other hand, “A study to determine the effects that ultraviolet light has on the ability of a fungus known as Saccharomyces cerevisiae to survive on YPD growth medium when incubated at 30°C” is an example of a title that is not succinct enough to be most effective. Plot the data as a scatter graph with a linear fit. Attach the graph to this report. (2 pts)(c) Use the linear formula to determine the percentage of surviving cells would be expected if 450 microJoules were used? (2 points)

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Laboratory 2: From Genes to Proteins to Behavior

CHARACTERISTICS OF RED YEAST:CAN THEY BE OBTAINED BY TREATMENT OF

WHITE YEAST WITH UV LIGHT?DO THEY HAVE SPECIAL NUTRITIONAL REQUIREMENTS?

Last week you treated white yeast cells with UV light. It was suggested that UV light can alter DNA.This week you will examine your UV-treated plates to see if any of the yeast cells changed their characteristics as a result of the UV treatment. We are particularly interested in any yeast that changed in color from white to red. Our working hypothesis will be that these and other changes occurred because UV light altered the DNA of the white yeast. The rest of the labs in the From Gene to Protein to Behavior project are designed to test this hypothesis.

Our focus this week will be in identifying and purifying colonies that are altered in appearance by theUV treatment. You should also notice more general effects of UV light on the yeast culture – do you notice any overall trends that are related to the dose of UV light received? If so, what might be the explanation for these trends?

Some students in the class will probably obtain red-colored colonies on their UV-treated plates. We are going to focus our future studies on these red yeast. Therefore, our second mission this week will be to begin studying previously identified strains of red yeast. You will use the streaking technique you learned last week investigate the growth characteristics of red and white yeasts. This plate test will address the question of whether red yeast and white yeast have different nutritional requirements. To generalize, we can ask whether there is a single difference between the two types of yeast (the color difference) or whether other differences exist, and if so, whether these differences are correlated. We will provide you with three yeast (Saccharomyces cerevisiae) strains (HAO, HB1, HB2) that are growing on rich medium, medium that has all the nutrients that the cells could possibly need. You will test whether these strains can grow on minimal medium, or medium that has limited types of nutrients.

In particular, you need to determine if the strains can grow without added adenine, a component ofDNA. Some strains can make their own adenine, but others cannot and therefore need to “eat” it from their environment before they can make their DNA.

Procedure:

1. Evaluating the effects of UV light on white yeast: Examine your petri plates from last laboratory period. Record in your lab notebook the number of colonies on each UV-treated plate and on the untreated control plates. In some cases the number of colonies will be so great that you can only estimate the number. To do this, count the number of colonies in one square centimeter of the plate and multiply by 63.6 (this is the total surface area of the plate in cm2). To keep track of which colonies have been counted, it is helpful to put a small dot of sharpie marker on the bottom of the plate over each colony as you count it.

Later in the semester, you will need these numbers, as well as notes about the other observations

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you will make regarding this project, for your lab report. Make sure you write your notes in a safe place and that EACH person in the lab group has ALL of the information every week.

You also will need to make a correction when comparing the number of colonies on your UV treated plates to that of the untreated plates. Remember, when you made your control plates you diluted your culture before plating it. You need to determine what was the dilution factor in this case.

Here’s an example of how dilution factor is calculated: if you had diluted 0.5 ml into 9.5 ml of water, then you would have diluted your 0.5 ml culture to make 10 ml of dilute culture. Your dilution factor would be 10 ÷ 0.5 = 20 fold dilution. Calculate the dilution factor for the dilution you used on your control plates.

Now you need to use this dilution factor in comparing the number of colonies on the UV-treated plates to your control plates. How do you think you can use the dilution factor in making this comparison?

Also record any differences you see between colonies on plates that had been subjected to UV light and colonies on plates that had not. Are there any examples of colonies that look unlike all other colonies (for example, red rather than the typical white color, have strange colony morphology, etc.)?

2. Streak plates to test nutritional requirements of red vs. white yeast: In addition to using the streak plate technique to purify colonies, it can also be used to place yeast on different growth media in testing their nutritional requirements. Obtain a minimal medium plate and a “mimimal plus adenine” plate. Using a marker on the bottom of the plates, draw lines dividing each plate into thirds (like a Mercedes hood ornament!) and label the three areas HAO, HB1, HB2. Also write on each plate what type it is (minimal, min plus adenine).

You are now going to streak three strains onto each of the two plates. In order to make the streaks fit into their sections of the plate, you will use a slightly different streaking pattern than is usually used in microbiology. Obtain plates (previously prepared by the instructors) with the yeast strains HAO, HB1, and HB2. Obtain a sterile plastic loop from your professor. Touch the loop to a single colony of HAO. Don’t scrape up a large amount of the colony – a barely visible or invisible amount is what you want. Remember cells are too small to see, so there are thousands of cells clinging to your loop, even though you can’t see them. If you pick up too many cells it will be difficult to interpret the experiment later. Touch the loop to the minimal medium plate and make a small zigzag streak across the third of the plate that you marked “HAO.”

Repeat this process for the HB1 and HB2 strains, making sure to use a new loop before touching each strain. Now, using the same procedure, streak all three strains on to the “minimal plus adenine” plate.

3. Streak plates to isolate red colonies from UV-treated plates: If you obtained any red colonies on your UV-treated plates, obtain one YPD plate to use in purifying the colony. Lightly touch a sterile loop to your red colony. Use the streak plate sterile technique that your instructor will show you in class, and the new YPD plate to purify the colony. If you succeed, you can use your pure red colonies next lab period to test your red strain’s nutritional requirements. If you found more than one red colony, please pass the extras on to lab groups that did not obtain a red colony. Don’t worry if you did not get to do this step; we will give you a red strain to work with in future weeks.

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Lab Assignment 2Due at the start of laboratory 315 points

Answer the following questions.(2 pts) 1. Which two sterile techniques described in this manual allow you to separate cells in a culture into individual colonies? Explain why these techniques can accomplish this separation.

3 Consider the following results:

(1 pt) (a) Can strain 1 make its own adenine?

(1 pt) (b) Can strain 2 make its own adenine?

(2 pts) (c) Explain how you arrived at your answers for (a) and (b).

(2 pts) (d) What situations might explain the results obtained with strain 3? The best answer will provide two different hypotheses. Consider possible technical problems and possible nutritional requirements of the strain in formulating your hypotheses. Propose a nutritional test experiment that could differentiate between the two possibilities.

(2 pts) 4. In your lab report, you will be asked to compare the number of colonies on your UV-treated plates to your control plates. How do you think you can use the dilution factor in making this comparison?

(2 pts) 5. In the following weeks you may be asked to interconvert milliliters (ml = 10-3 liters) and microliters (μl = 10-6 liters). How many microliters are in 0.3 ml? How many milliliters are in 250 μl? (please show your work)

(2 pts) 6. Look at the flowchart entitled “Simplified adenine synthesis pathway” in the lab manual’s table of contents. The flowchart shows a small portion of the steps needed to make adenine. Three genes encode three proteins (enzymes) – for example, gene A contains a code telling the cell how to build protein A. Each protein does a job. For example, protein A converts a colorless chemical, represented by the open shape, into a red chemical, represented by the filled-in bullet shape. The proteins must do their jobs in the order indicated – protein B cannot do its work until protein A has done its job. A student from last year’s class claims that if a cell contains an altered version of gene A that encodes a non-functional version of protein A, then the cell will remain colorless (white). However, if a cell contains an altered version of gene B that encodes a non-functional version of protein B, then the cell would turn red. Does this seem possible? Explain your answer.

(1 pt) 7. Would an alteration in gene C leading to a non-functional protein C result in red or colorless/white yeast? Briefly explain.

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Laboratory 3: From Genes to Proteins to Behavior

PREPARATION OF DNA FROM A RED YEAST STRAIN

In the first lab you subjected a white yeast strain to ultraviolet light. After the UV treatment some of you observed a red colony amidst the thousands of white colonies. In other studies you have seen that red yeast and white yeast strains isolated previously have other differences in addition to color. The two types of strains also showed a difference in their growth properties. The red strains were unable to grow unless adenine was provided to the agar nutrients while the white strain was able to grow in the absence of adenine. Today you will check any red strains you or your classmates isolated to see if they too require adenine for growth.

Because UV treatment is known to alter DNA, both the difference in colony color and nutritional requirements could be due to an alteration in the DNA. A particular DNA change could convert a white, adenine-synthesizing strain into a red, adenine requiring strain. This is our working hypothesis. If our hypothesis is correct then we should be able to identify a difference in the DNA when we compare the DNA of the red and white types. To study the DNA we will separate it from all of the other cellular constituents. This week you will prepare a purified sample of DNA from cells that formed red colonies after being subjected to UV light. DNA from cells that form white colonies has already been analyzed in an international multi-million dollar collaboration similar to the much-touted human genome project. We can therefore compare the results from the DNA analysis from the cells that formed red colonies after UV exposure to the DNA from the cells that form white colonies.

We will extract DNA from yeast using the MasterPure Yeast DNA Purification kit. It enables efficient, high-yield purification of high-molecular-weight DNA from yeast and other fungi. The protocol involves nonenzymatic cell lysis at 65°C, followed by removal of protein by precipitation, and nucleic acid precipitation and resuspension. 

Procedure

1. Scrape a single yeast colony or yeast mass (2mm in diameter) from an agar plate and transfer to a microcentrifuge tube containing 300 l of Yeast Cell Lysis Solution (provided).

2. Suspend the cells by pipetting the cells repeatedly using a P1000 set to 150 L. 3. Incubate the suspended cells at 65C for 15 minutes. 4. Place the sample on ice for 5 minutes. Add 150 L of MPC Protein Precipitation reagent and mix by

pipetting up and down for 10 seconds. 5. Pellet cellular debris by centrifugation in a microcentrifuge for 10 minutes at 13000 rpm. 6. Transfer the supernatant to a clean microcentrifuge tube and add 500 L of isoproponal. Mix

thoroughly by inversion.7. Pellet the DNA by centrifugation in a microcentrifuge for 10 minutes at 13000 rpm. 8. Remove the supernatant by pipetting and discard it. 9. Wash the pellet containing the DNA with 500 L of 70% ethanol. 10. Centrifuge briefly for 1 minute at 13000 rpm. Carefully remove the ethanol by pipetting and discard

it. Remove all traces of ethanol.11. Suspend the DNA pellet in 35 L of sterile water. Use this as your template Yeast DNA in your PCR

reactions.

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Laboratory 4: From Genes to Proteins to Behavior

AMPLIFYING A SEGMENT OF THE ADE1 AND ADE2 GENES BY THE POLYMERASE CHAIN REACTION

&CHARACTERISTICS OF THE ADE1 AND ADE2 GENES FROM THE

WHITE YEAST

For the past several weeks you have been working with two strains of yeast, one whose cells form red colonies and one whose cells form white colonies. In the first week of lab you subjected the yeast that gives rise to white colonies to UV light. On rare occasions this treatment resulted in colonies that were red. The lab work at this point forward is aimed at defining the underlying difference between the red and the white types of yeast that explains the difference in color. And does this difference explain the difference in nutritional requirements?

Because UV light is known to mutate (change) DNA, we are focusing our attention on comparing the DNA from the red and white yeast. In order to analyze the DNA we needed to separate it from all of the other cellular constituents (parts). This is what you did last week in the laboratory.

There are two objectives in today’s laboratory period. The first objective is to make many copies of two segments of DNA (genes; ADE1 and ADE2). The technique known as polymerase chain reaction (PCR) will be used. This technique will allow us to generate enough copies of the DNA to allow for the determination of the sequence of bases of the genes ADE1 and ADE2. You will use a technique called agarose gel electrophoresis in order to determine whether there is DNA in the final solution you recovered last week.

PCR amplification of ADE1 and ADE2 from Red Yeast

Polymerase Chain Reaction (PCR) is an in vitro (in test tube) method to replicate DNA. It bears some basic resemblance to the way DNA is replicated inside the cell but also differs in some fundamental ways. The power of PCR technology lies in the ability to start with tiny quantities of DNA material that are essentially undetectable and selectively amplify a region of interest into such amounts that the amplified DNA is simple to detect.

The PCR process involves three steps in repeated cycles of varying temperatures. The cycling allows replication of the DNA by 1) denaturation of double stranded DNA template (but not the thermostable polymerase) at 95°C, 2) annealing of the short complementary DNA oligonucleotides (primers) to the single-stranded template at 55°C, then 3) extension of the DNA strand from the primers by the polymerase at 72°C. The thin-walled PCR tubes and the rapid temperature change of the thermocycler allow these cycles to occur very rapidly. One copy of the double-stranded template can be amplified to more than a billion (230 >109) with 30 cycles in 2-3 hours in a 25 µL reaction (Fig 1 from Brooker).

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The strains of yeast with red color have been associated with a defect in the ability to make adenine. Adenine is chemically related to several molecules that have a red color, although adenine itself does not. Normally these molecules that have a red color are converted to adenine by two different enzymes. Two genes, ADE1 and ADE2, encode these enzymes. A defect in either one of these enzymes can lead to the accumulation of one of these molecules that have red color, thus the red color yeast. In this lab you will amplify DNA from the ADE1 gene and the ADE2 gene.

The following components are required to perform a PCR:

1. Source of DNA to be amplified: This starting material may be purified DNA or can be small amounts of tissue or cells. In forensic medicine, one is often forced to use small bits of tissue left at the crime scene. You will use the DNA that you isolated from a single colony of red yeast.

2. A pair of short, single stranded DNA molecules called oligonucleotides that will be used to prime the replication reaction. Because of their function these oligonucleotides are also called primers. In this particular case, you will be given primers of the appropriate base sequence to amplify the ADE1 gene and the ADE2 gene.

3. Thermostable DNA Polymerase enzyme: DNA polymerases are enzymes that have the ability to incorporate deoxyribonucleotides into a growing chain of DNA. For PCR, it is essential that the

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enzyme be heat stable (thermostable). Heat stable enzymes are generally found in bacterial sources that grow in very harsh environments. The particular enzyme used most often is called Taq polymerase. Thermus aquaticus grows in and around hot geysers.

4. deoxynucleotide triphosphates (dNTP’s): These are the building blocks of DNA and are the molecules that the polymerase enzyme incorporates into the growing DNA chain.

5. Buffer: This will make sure that the pH and ionic requirements for the DNA polymerase are met.

You will be using Ready-To-Go PCR beads to setup your PCR reactions. This bead has the dNTP’s, the DNA polymerase, and the buffer. Traditionally there would be three separate solutions to pipet (dNTPs, DNA polymerase and the buffer).

Procedure

1. Obtain from your instructor a. 2 tubes, each containing a Ready-To-Go PCR bead b. Ade1 primer mix (contains the forward and reverse primer to amplify ADE1)c. Ade2 primer mix (contains the forward and reverse primer to amplify ADE2)

2. Label 1st tube as Ade1 and the 2nd tube as Ade2. Add your group # to both tubes so that you can identify your tubes.

3. To the Ade1 tube add the following in the order listed below. Make sure to use a clean tip for each reagent.

Ade1 primer mix 12.5 l Water 7.5 l Yeast DNA 5.0 l

25.0 l (Total volume)

4. To the Ade2 tube add the following in the order listed below. Make sure to use a clean tip for each reagent.

Ade2 primer mix 12.5 l Water 7.5 l Yeast DNA 5.0 l

25.0 l (Total volume)

5. Close the tubes and gently flick the tubes with your finger until the bead is completely dissolved.

6. Place the tubes in a microcentrifuge and spin for 1 min at 3000 rpm.

7. Place the tubes in the thermocycler. It will be started when all the groups have placed their tubes into the thermocycler.

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(Note that to successfully complete the second objective you have to successfully complete the DNA preparation). The analysis of the base sequence of the two genes from the red yeast will be the goal of the next laboratory period. The third objective of this lab is to learn about the characteristics of theADE1 and ADE2 genes from the white yeast; this information is accessible in a public database.

Laboratory 5: From Genes to Proteins to Behavior

AGAROSE GEL ELECTROPHORESIS AND CHARACTERISTICS OF THE ADE1 AND ADE2 GENES FROM

THE RED YEAST

I. AGAROSE GEL ELECTROPHORESIS

An introduction to gel electrophoresis can be found in your textbook. Below is a brief description.DNA fragments differing in length can be detected and separated by gel electrophoresis. A mixture of negatively charged DNA fragments is placed in a porous gel and placed in the presence of an electric field. The negative charge on each DNA fragment is proportional to its length because each phosphate group in the chain in negatively charged. As the fragments move through the electric field, attracted by the positively charged electrode, the gel resists their movement but with different intensity; the longer fragments are retarded more strongly than the smaller fragments. Therefore the smaller fragments move the fastest and travel the farthest across the gel. When the electric field is removed, each DNA fragment will have stopped at a different point on the gel. Each visible "pile" of DNA is called a band, and the path that the DNA travels in is called a lane. If desired, the separated fragments can be removed from the gel in their pure form.

Different samples may be “run” side by side in different “lanes” in the electric field. Conventionally, and in today’s laboratory experiment, DNA fragments of known length are run next to experimental mixtures to provide a size reference.

The DNA on the gel can be visualized by staining the gel with a fluorescent dye and subsequently placing the gel under ultraviolet light. This dye is known as EZ Vision, and will be added to your sample just before you load it in the well.

Recall that last lab you amplified ADE1 and ADE2 gene segments from yeast cell DNA. We have no verification however, that you were successful. Everyone in the lab has an eppendorf tube with a clear liquid solution. Is there any DNA in that solution? If so, is it the expected length? Today we will use agarose gel electrophoresis to determine if your tube does indeed have DNA and assess whether the DNA is the expected size.

Methods1. Obtain your amplified yeast DNA sample from the instructor.2. Using a pipetman (P20) withdraw 5 microliters of your DNA sample and pipet it into an

eppendorf tube.3. Using the P2, add 1 microliter of EZ Vision loading dye to your sample. Close the lid and mix the

dye and your DNA sample together by pipetting gently up and down.

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The loading dye serves three purposes. First, it gives the sample a density greater than water. This means when you add your sample to an aqueous medium it will sink. Second, the migration of the dye solution will give an indication of how long the sample should remain in the electric field. Finally, EZ Vision contains a fluorescent dye that binds to the DNA and that will allow us to visualize it using ultraviolet light.

4. The agarose gel may have been prepared in advance. If it has been, take your sample mixed with dye to the electrophoresis chamber. The gel will be loaded as a group effort. Take careful notes so that you know which lane has your sample and which samples were loaded in the other lanes. Your instructor will also load a DNA size standard at this time. After all of the samples are loaded on the gel, a current will be passed between the two electrodes. This will start the process of electrophoresis. After about 45 minutes the samples should be ready to visualize.

II. CHARACTERISTICS OF THE ADE1 AND ADE2 GENES FROM THE WHITE YEAST

Mining a database for information about the ADE1 and ADE2 genes and their corresponding proteins

A gene is a segment of DNA that codes for a protein. A unicellular organism such as S. cerevisiae has several thousand genes. In the case of this yeast, the sequence of bases that make up all of the genes is known. This is a Herculean task and has therefore only been done for the normal white type and not for the red type of yeast. Today’s objective is to locate in a database the two genes of the white form of yeast that we are interested in ADE1 and ADE2. Since the sequence of the bases is already known, we should be able to find this information on the internet. We will also find the sequence of amino acids that make up the protein that is encoded by each of the two genes. Remember that the long-term objective of the project is to determine if there has been a change in the sequence of bases in either of two genes, ADE1 and ADE2, in the yeast that form the red colonies compared to those that form the white colonies. This objective will be accomplished by next week.

For today, each lab pair will divide the tasks. One person will work with the ADE1 gene while the other will do the same for ADE2. For those that are working on the ADE2 gene, skip down to the heading: "Finding the ADE2 gene in the database".

Finding the ADE1 gene in the database.

1. Open the internet browser and type in the address http://www.yeastgenome.org

This is the portal to the Saccharomyces Genome Database (SGD), the database that will have the sequence of bases for all six thousand genes. But note that this database is only for the white type of yeast.

2. In the box that is labeled “Search our Site” type “ADE1” and click on the “Submit” button.

3. On the screen that follows, you should see a table with two choices. Each gene in the database has a Gene name like ADE1 and another identifier (Systematic Name) that indicates (in a nonobvious way) the location (“address”) of the gene on a particular chromosome. Click on the YAR105W in the table.

4. The screen that now shows is fairly complicated. The title for the page is ADE1/ YAR105W Summary. On

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the left half of the screen there is a column of buttons. Look for the heading “Sequence Information” and then “Retrieve sequences.” Click and hold on “ORF Genomic DNA.” You will see five options to select from on the drop-down menu. Choose “coding sequence” and then click on the button to the right “Get Sequence”. Like all structural genes, the ADE1 gene codes for a protein. By selecting this option the display now shows the DNA sequence necessary to code for this protein.

The current screen shows the bases that make up the ADE1 gene in FASTA format. There are other ways of arranging this same data, one of which turns out to be much easier to determine features such as length (in number of bases). This format is known as GCG. To see the GCG format click on “GCG” next to the phrase “Other formats available”. Notice that this format has numbers that signify the number of bases from the first base at the start of each row. The bases are presented in groups of 10 with a space between each group provided for your viewing pleasure. Just prior to the presentation of the base sequence is a reference to the original scientific manuscript that first identified and reported the base sequence of ADE1.

The bases that are shown represent the sequence of ONE strand of the DNA. For the questions below you do NOT have to figure out the sequence of the RNA. Recall, one strand of the DNA in a gene is used as a template to form an mRNA, and then this mRNA is used in translation as instructions for building a protein. The DNA sequence that you are viewing is the coding strand of the ADE1 gene. Like the mRNA, the coding strand is complementary to the template strand of the DNA. Thus, the mRNA sequence and the sequence of the coding strand are identical, except for the substitution of a ‘U’ in mRNA every place there is a ‘T’ in DNA.

example:

DNA 5’ ATGGATTCTAGA 3’ coding strand of DNA

3’ TACCTAAGATCT 5’ template strand of DNA

mRNA 5’ AUGGAUUCUAGA 3’ mRNA transcribed from the template strand of DNA

Note that this mRNA sequence is identical to the coding DNA strand, except for the replacement of T’s with U’s. Therefore, you can look up the encoded amino acids directly in a genetic code table simply by substituting U’s for T’s in the sequence that you are viewing.

What are the first three bases that are displayed? What special meaning do these three bases have?

How many nucleotides make up the ADE1 gene?

What are the last three nucleotides? What amino acid is coded by these three nucleotides?

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Use the chart in the handout or your textbooks to determine the sequence of the first ten amino acids. Write them in the space below using the single letter abbreviation for each amino acid. The single letter abbreviations will be provided.

___ ___ ___ ___ ___ ___ ___ ___ ___ ___

Obtaining the protein sequence encoded by the ADE1 gene

5. In principle you should be able to determine the entire sequence of amino acids of the protein encoded by the ADE1 gene. You have already determined the first 10 amino acids. It did not escape our attention how tedious that task was. This type of work is best done by a computer. So we will now find the entire sequence of amino acids encoded by the ADE1 gene but we will let the computer do the work.

Click on the “back” button twice on the internet browser to return to the original screen. Under the heading “Retrieve sequence” select the option “ORF Translation” and then click on “Get Sequence”. (ORF is an abbreviation for open reading frame. An ORF is a sequence of DNA that can code for a protein.) Note that the screen that is generated has characters of almost all letters in the alphabet. This string of characters on the display represent single letter abbreviations of amino acids. Check to see if the first ten amino acids that you predicted in step 3 are consistent with what is being displayed here. In order to see the same data displayed in a different manner, click on “GCG” next to the phrase “Other formats available.”

How many amino acids are connected to make the protein encoded by the ADE1 gene?

How many nucleotides were required to code for this many amino acids?

6. Once more click on the “Back” button twice on the internet browser to return to the ADE1 main page. On the left side of the page examine the information under “Description”.

Based on the information provided, is it possible that a change (mutation) in the ADE1 sequence could lead to the red color property?

Click on the “Phenotype” tab at the top of the table and examine the heading “Classical Genetics”. Note that a comment similar to the one under “Description” is here. In addition, note the additional comment in this section “auxotroph”. The term “auxotroph” means that something must be added in order for the organism to grow.

Is this information consistent with the hypothesis we are testing and any observations that you have made?

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7. You will now collect a bit more information about the ADE1 gene and the protein encoded by the ADE1 gene. The top of the page gives a description of some of the properties of the gene and the protein. Under the heading “Description” is the name of the enzyme encoded by ADE1. It is certainly a mouthful--- N-succinyl-5-aminoimidazole-4-carboxamide ribotide synthetase---but note the suffix "-ase" that always signifies that the term is describing an enzyme. We will come back to this name in the computer lab next week when we find a human protein that has the same name.

Next look at the entry under "Biological Process" on the left side of the page. Here the entry states "purine nucleobase metabolic process". Metabolism is the process of construction or destruction of biological molecules, in this particular case purine bases.

Do you have any experimental evidence (if not, how could you attempt to gather such evidence) that cells that give rise to red colonies are also affected in purine base metabolism?

Finding the ADE2 gene in the database.

1. Open the internet browser and type in the address http://www.yeastgenome.org

This is the portal to the Saccharomyces Genome Database (SGD), the database that will have the sequence of bases for all six thousand genes. But note that this database is only for the white type of yeast.

2. In the box that is labeled “Search our Site” type “ADE2” and click on the “Submit” button.

3. On the screen that follows, you should see a table with two choices. Each gene in the database has a Gene name like ADE2 and another identifier (Systematic Name) that indicates (in a nonobvious way) the location (“address”) of the gene on a particular chromosome. Click on the YOR128C in the table.

4. The screen that now shows is fairly complicated. The title for the page is ADE2/YOR128C Summary. On the left half of the screen there is a column of buttons. Look for the heading “Sequence Information” and then “Retrieve sequences.” A Click and hold on “ORF Genomic DNA.” You will see five options to select from on the drop-down menu. Choose “coding sequence” and then click on the button to the right

“Get Sequence”. Like all structural genes, the ADE2 gene codes for a protein. By selecting this option the display now shows the DNA sequence necessary to code for this protein.

The current screen shows the bases that make up the ADE2 gene in FASTA format. There are other ways of arranging this same data, one of which turns out to be much easier to determine features such as length (in number of bases). This format is known as GCG. To see the GCG format click on “GCG” next to the phrase “Other formats available”. Notice that this format has numbers that signify the number of bases from the first base at the start of each row. The bases are presented in groups of 10 with a space between each group provided for your viewing pleasure. Just prior to the presentation of the base sequence is a reference to the original scientific manuscript that first identified and reported the base sequence of ADE2.

The bases that are shown represent the sequence of ONE strand of the DNA. For the questions below you do NOT have to figure out the sequence of the RNA. Recall, one strand of the DNA in a gene is used as a template

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to form an mRNA, and then this mRNA is used in translation as instructions for building a protein. The DNA sequence that you are viewing is the coding strand of the ADE2 gene. Like the mRNA, the coding strand is complementary to the template strand of the DNA. Thus, the mRNA sequence and the sequence of the coding strand are identical, except for the substitution of a ‘U’ in mRNA every place there is a ‘T’ in DNA.

example:

DNA 5’ ATGGATTCTAGA 3’ coding strand of DNA

3’ TACCTAAGATCT 5’ template strand of DNA

mRNA 5’ AUGGAUUCUAGA 3’ mRNA transcribed from the template strand of DNA

Note that this mRNA sequence is identical to the coding DNA strand, except for the replacement of T’s with U’s. Therefore, you can look up the encoded amino acids directly in a genetic code table simply by substituting U’s for T’s in the sequence that you are viewing.

What are the first three bases that are displayed? What special meaning do these three bases have?

How many nucleotides make up the ADE2 gene?

What are the last three nucleotides? What amino acid is coded by these three nucleotides?

Use the chart in the handout or your textbooks to determine the sequence of the first ten amino acids. Write them in the space below using the single letter abbreviation for each amino acid. The single letter abbreviations will be provided.

___ ___ ___ ___ ___ ___ ___ ___ ___ ___

Obtaining the protein sequence encoded by the ADE2 gene

5. In principle you should be able to determine the entire sequence of amino acids of the protein encoded by

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the ADE2 gene. You have already determined the first 10 amino acids. It did not escape our attention how tedious that task was. This type of work is best done by a computer. So we will now find the entire sequence of amino acids encoded by the ADE2 gene but we will let the computer do the work.

Click on the “back” button twice on the internet browser to return to the original screen. Under the heading “Retrieve sequence” select the option “ORF Translation” and then click on “Get Sequence”. (ORF is an abbreviation for open reading frame. An ORF is a sequence of DNA that can code for a protein.) Note that the screen that is generated has characters of almost all letters in the alphabet. This string of characters on the display represent single letter abbreviations of amino acids. Check to see if the first ten amino acids that you predicted in step 3 are consistent with what is being displayed here. In order to see the same data displayed in a different manner, click on “GCG” next to the phrase “Other formats available.”

How many amino acids are connected to make the protein encoded by the ADE2 gene?

How many nucleotides were required to code for this many amino acids?

6. Once more click on the “Back” button twice on the internet browser to return to the ADE2 main page. On the left side of the page examine the information under “Description”.

Based on the information provided in the “Summary Paragraph”, is it possible that a change (mutation) in the ADE2 sequence could lead to the red color property?

Click on the “Phenotype” tab at the top of the table and examine the heading “Classical Genetics”. Note that a comment similar to the one under “Description” is here. In addition, note the additional comment in this section “auxotroph”. The term “auxotroph” means that something must be added in order for the organism to grow.

Is this information consistent with the hypothesis we are testing and any observations that you have made?

7. You will now collect a bit more information about the ADE2 gene and the protein encoded by the ADE2 gene. The top of the page gives a description of some of the properties of the gene and the protein. Under the heading “Description” is the name of the enzyme encoded by ADE2. It is certainly a mouthful--- phosphoribosylaminoimidazole carboxylase---but note the suffix "-ase" that always signifies that the term is describing an enzyme. We will come back to this name in the computer lab next week when we find a human protein that has the same name.

Next look at the entry under "Biological Process" on the left side of the page. Here the entry states "purine nucleobase metabolic process". Metabolism is the process of construction or destruction of biological molecules, in this particular case purine bases.

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Do you have any experimental evidence (if not, how could you attempt to gather such evidence) that cells that give rise to red colonies are also affected in purine base metabolism?

II. ANALYSIS OF THE BASE SEQUENCE IN THE DNA OF THE ADE1 AND ADE2 GENES FROM RED YEAST

In the first laboratory period you examined the effect of UV light on S. cerevisiae. Many of you were able to note that this treatment resulted in a number of different types of variants, including colonies that appeared red. That the red colony types arose from the white suggests that the two are highly related (highly similar). Are there any other differences between red and white strains in addition to the difference in color? You tested the growth properties of the red and white yeast under two conditions, with adenine provided in the agar and without adenine provided in the agar. The two types of yeast showed a difference with respect to their growth. Thus the two types of yeast, although presumably highly similar, seem to differ with respect to two characters, colony color and nutritional requirements.

So in those colonies that were red that came from a white progenitor, what changed? What is the underlying difference that leads to the difference in colony color?

In the third week of laboratory you prepared DNA from the red type of yeast. We are testing the hypothesis that the difference between the red and the white yeast is a difference in one of two genes. The names of the two genes are ADE1 and ADE2. Genes are segments of DNA and therefore in order to test this hypothesis we need to obtain and characterize the DNA from the red and the white yeast. We will examine ADE1 and ADE2 genes (there are a total of 6000 genes in the yeast) because in previous cases red colonies were associated with changes in one of these two genes. You noted this in the computer work you did last week in lab. By the end of the third laboratory period you obtained DNA from the red type of yeast.

(There are no set rules for naming of genes but it is typical that the name of a gene reflects the function of the protein encoded by the gene. Recall that in the first week you showed that the red version of the yeast required adenine in order to grow while the white version of yeast did not require adenine to grow. Previous researchers have shown that in versions of yeast where cells are red and adenine is required to grow there is a change in one of two genes. Because changes in each of these genes results in a requirement for a d e nine the genes were named ADE1 and ADE2.)

In the fourth week of laboratory you performed the polymerase chain reaction (PCR) on the DNA from the red yeast strain. This DNA contains the entire repertoire of 6000 yeast genes. Remember, the hypothesis that we are testing is that the difference between the standard (wild-type) white yeast strain and the red yeast strain is due to differences in the base sequence of either the ADE1 or ADE2 genes. The PCR was done in order to restrict our analysis to only two of the 6000 genes and to obtain sufficient quantities of the DNA of the two genes (ADE1 and ADE2) so that the base sequence of each of these two genes could be determined. Also remember that we already know the sequence of the two genes in the standard white type of yeast. This is what you found on the computer last week.We have taken those DNA molecules generated by PCR and sent them to a DNA sequencing facility. Such facilities have the equipment to determine the base sequence in the DNA.

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The data have come back from the sequencing facility and we will spend today’s laboratory period analyzing and interpreting data through the use of computer software. The data analyzed will be from only one of the red mutant yeast strains. These data will then be compared to data you obtained last week on the standard (wild-type) white strain. Will there be any differences between the data we have collected from the red strain when compared to the sequence data of the white strain? If so, will differences be detected in both the ADE1 and ADE2 genes?Sequence Analysis of the Red Yeast Mutant

The technology of DNA sequencing has revolutionized biology. This technique is described in your textbook; you will also work through some critical thinking questions to better understand the technique. In addition, a short animation describing how DNA is sequenced will be shown. This will give you an indication of how the following sequence data (see Step 1) was obtained. The computer files that you will need will be provided by your instructor. The computer program you will need (4Peaks) is on the PowerBook notebook computers. Note that each question that is both underlined and in italics will be questions for the lab assignment.

During today’s laboratory period you will use the computer to analyze the ADE1 and ADE2 genes from the red yeast and compare the data with the ADE1 and ADE2 genes of the standard white yeast. One person in each lab group will study the ADE1 gene while the other person will study the ADE2 gene. You will then share your results with your lab partner. If you will be studying ADE1, then start reading below. If you will be studying ADE2, skip about eight pages and start on the page the top of which reads “Below are the objectives and procedures for studying the ADE2 gene.”

Below are the objectives and procedures for studying the ADE1 gene.• Examining the primary data of the ADE1 gene• Converting the ADE1 sequence to that on the complementary DNA strand.• Comparing the nucleotide sequence of the ADE1 gene from the red yeast to the ADE1 gene

from the white yeast• Determining the amino acid sequence of the protein encoded by the ADE1 gene• Comparing this amino acid sequence to the sequence from the standard (wild-type) white

strain• Examining the database for a human gene that encodes a (weakly) similar protein that

carries out the same function

1) Examining the primary data of the ADE1 geneOpen up the file labeled “ADE1r.ab1” using the program “4Peaks”. To do this double-click on the “4Peaks” icon, and under the pulldown heading of “File” select “Open”. Then select the file name “ADE1r.ab1”. This computer file contains the sequence data we obtained from the PCR product generated with the ADE1 primer set in the previous lab session. This is the sequence of bases corresponding to the ADE1 gene from the red yeast strain. At this point you should see peaks offour different colors. There is also an X-axis that consists of a time unit of measurement, anucleotide number, and a nucleotide (designated A, C, G, or T). There are a few things to note:

i) at many points there is only one peak and that the color of the peak determines the nucleotide indicated on the X-axis.

ii) at some points, there is more than one peak (see bases near the beginning, for example).This is due to technical problems and makes it difficult or impossible to predict the base that is at that position on the DNA strand. This is why there is occasionally a “N” instead of an A, C, G, or T. “N” designates that the identity of the base at this position was not clearly determined.

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iii) it is technically more difficult to determine the base sequence near the beginning. It is also technically difficult to determine the base sequence much more than about 500 nucleotides from the start. Therefore we will limit our analysis to the base sequence between positions115 and 540.

Highlight (select) the characters between position 115 and 540 (from ACACCG to TTTTTACG) and then select "Crop" from the drop-down menu in the lower left hand corner of the window (the ‘gear’ icon). The sequence that remains in the 4Peaks window is now just the DNA between bases 115 and 540. Use “Save as” to rename this cropped file.

2) Converting the ADE1 sequence to that on the complementary DNA strand. Remember that DNA has two strands of nucleotides that run in opposite directions and are complementary. The sequence you have shows only one strand of the DNA, but because of the strict base pairing rules you can derive what the other strand must look like. For example, if the sequence of one strand is5’ CCACGT 3’ then the sequence of the complementary strand must be 5’ ACGTGG 3’. As luck would have it, the sequence you have received is actually the sequence of the gene on the template strand, read from the end of the gene to the beginning. In other words, the sequence is in reverse and complementary to the mRNA sequence that would be used to make a protein. In order to make the analysis of the ADE1 gene easier, we need to turn the sequence around so that it reads like the other DNA strand (called the “non-template strand”). We will use 4Peaks to make this change. We could do this manually, but it would be very tedious to write out the complement of hundreds of bases.

To convert the base sequence into the complementary base sequence, select “Flip” from the drop-down menu in the lower left hand corner of the window (the ‘gear’ icon). Note that the new sequence that is generated is simply the sequence of the second strand of the original DNA displayed 5’-> 3’. Now select this new sequence and select “copy” from the Edit menu.

3) Comparing the sequence of the ADE1 gene from the red yeast to the sequence of the ADE1 gene from the white yeast. In the following steps you will search SGD, the database that has the DNA sequence of all genes from the standard (wild-type) white yeast that you used last week, tosee if any have the same, or similar, sequence of bases as the sequence you have determined for theADE1 gene from the red yeast. To prepare for the comparison, copy the sequence by selecting “Copy Sequence” under the “Edit” menu.

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a) Accessing the database via the internetOpen a web browser and type in the following address http://www . y ea s tg e nom e . o r g

At this website, the entire base sequence of the standard (wild-type) white strain ofSaccharomyces cerevisiae is stored. This represents about 10 million base pairs. Nested somewhere in this 10 million base pairs is the sequence for the ADE2 gene (only about 2000 base pairs long). At the top of this web page within the purple bar, scroll your mouse over “Analyze” and select “B L A S T ”. This program will search the entire sequence of bases of the Saccharomyces cerevisiae in attempt to find some sequence that is either identical or similar to the “query” sequence (BLAST is an acronym for Basic L ocal Alignment S equencing Tool). The query sequence is that which you type or paste into the window, as in the following step. In this case, the query sequence is the ADE1 base sequence from the red yeast strain.

Under the heading “Choose the Appropriate BLAST Program” make sure that BLASTn is chosen. This is the DNA/RNA database. N stands for nucleotide).

Under the heading “Choose One or More Sequence Datasets” you should select “Open Reading Frames (DNA or Protein)”.

b) Searching the DNA database: Now place the cursor into the window above which is written “Type or Paste a Query Sequence” and paste the sequence that you copied in step 1. Use either the Paste command from the Edit menu or hold down the command and “V” keys to paste. Next select the button “Run WU-BLAST”. The query sequence (the ADE1 sequence from the red yeast strain) is now being compared to all 10 million base pairs in the database of the standard (wild-type) white yeast strain to find the sequence or sequences that are most similar.

c) Matches to the query DNA sequence: After a short time you will see the results of this search although it will be difficult to interpret. Near the top of the page you will see what looks like a number line. This line represents the sequence you put in, with the numbers indicating the nucleotide number. Under this numbered line you will see one or more colored thick lines (actually they are arrows). Each thick line represents a match to the query sequence. In this case there is one match (there is one line), represented by a thick green arrow. In order to see how closely the two base sequences match scroll down; you should see lines of base sequence (strings of A, C, G, and T) and hash-marks between them. Note the similarity in the DNA sequence between the query (the ADE1 gene of the red yeast) and the DNA sequence found in the database (labeled "Sbjct"; this is the ADE1 gene of the white [wild-type] yeast). Note that one of the identifiers of the subject sequence is ADE1 and that the description includes the name of an enzyme that we have seen before and the comment that the gene is required for the synthesis of adenine. The query sequence starts with base number (coordinate) 1 and goes to coordinate 426; the "Sbjct" sequence starts at coordinate 399 and goes to coordinate 824. Note that each line of sequence alternates between the query and the sbjct sequence. A hash-mark between the query and sbjct sequences indicates that the base sequence is identical between the query and the sbjct at that coordinate. The top portion of the screen will look like this:

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d) But is it an exact match? Recall that our hypothesis is that the difference between the white and red yeast strain is attributable to difference in the base sequence of the ADE1 or ADE2 genes. Such a difference could be fairly small so check carefully to see if the query sequence and the Sbjct sequence are exact matches.

i. Do y ou s e e any d i ff e r e n ce s ? A ny diff e r e n c e w ould m e an that the ba s e s e qu e n c e of the A D E 1 g e ne of the r e d ye a s t is diff e r e nt f r om the ba s e s e qu e n c e of the A D E 1 g e ne o f the w hite ye a s t. I f a diff e r e n c e w as ob s e r ve d, w hat t r e atm e nt of the ye a s t in w ee k 1 l e d to this di f f e r e n c e in the ba s e s e qu e n ce ? As a reminder, the base sequence in the database (the “Sbjct”) came from the standard (wild-type) white yeast.

Su mm a r y of S t e p s 1 - 3. At this point you have determined the sequence of bases that make up the ADE1 gene from the red yeast, searched a database and compared the sequence of bases of the ADE1 gene from the white yeast to the ADE1 gene of the red yeast. In the next step you will use the base sequence of the ADE1 gene from red yeast to determine the amino acid sequence of the protein encoded by the ADE1 gene from the red yeast. You will then compare the amino acid sequences from the red and white yeasts.

4) Determining the amino acid sequence of the Ade1 protein from the red yeast. In this section you will use the 4Peaks program to determine the amino acid sequence of the protein encoded by the ADE1 gene.

a) What’s in a DNA sequence? Recall our definition of a gene: a gene is a segment of DNA that codes for a protein. As was the case last week, the computer shows you the sequence of the non-template strand of DNA, the strand that looks like the mRNA, except for the presence of T’s instead of U’s. This strand can be directly analyzed using a genetic code table to decipher what amino acids are encoded. The DNA code is written in groups of three bases. Note that the “reading frame” of the sentence is quite important. For instance, if I wrote a sentence of three letter words in the midst of a random assortment of letters it would be nonsense unless you started in the correct reading frame. The three reading frames of the sentence TETHECATATETHEBIGRATTT would be:

Reading frame (1) TET HEC ATA TET HEB IGR ATT T(groups of three start at the first letter)

Reading frame (2) T ETH ECA TAT ETH EBI GRA TTT(groups of three start at the second letter)

Reading frame (3) TE THE CAT ATE THE BIG RAT TT(groups of three start at the third letter)

There is only one of three reading frames in which you will be able to decipher the meaning of

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this sentence. So in this example, unless you start reading in the “3rd” reading frame, this string of letters would have no meaning. Likewise, to decipher the amino acid sequence of a gene the information must be read in the correct “reading frame”. The 4Peaks program will examine each of the three possible reading frames to determine which one has a “sentence”. In this case a sentence means a long string of amino acids that are encoded without any “stop” codes. Recall that some 3- letter combinations represent a “stop”---that is, rather than an amino acid this stop signal results in the termination of translation. The codes for stop are TGA, TGG, and TAG.

b) What and where is an ORF? An “ORF” is an abbreviation for an open r eading frame and is metaphorically the same as finding a sentence in the midst of a number of letters strung together. The 4Peaks program has tools designed for this purpose. Return to 4Peaks. Click on the Show Translation icon in the lower right hand corner of the window. This operation will “decode” (translate) your DNA sequence into an amino acid sequence by analyzing the nucleotide sequence in groups of three, noting any cases of stop codes as asterisks (*). Click on the “Set Translation Frame” icon in the lower right hand corner, to select a different reading frame. An open reading frame (orf) will contain no asterisks. This means that this segment of DNA has the potential to code for a protein because there are no stops, just amino acids, encoded.

c) The amino acid sequence of the protein encoded by the ADE1 gene of the red yeast. If you do see an ORF this indicates the potential for this sequence of bases to encode a protein. This is the sequence of amino acids in the protein encoded by the red yeast ADE1 DNA that you sequenced. The amino acids are given in one-letter abbreviations. For example, the amino acid serine is abbreviated with the letter S and the amino acid leucine is abbreviated with the letter L. If you have succeeded then the last seven amino acids using the one letter abbreviation should be "TANKLNG".

5) Comparing the amino acid sequence encoded by the ADE1 gene from the red yeast to the amino acid sequence encoded by the ADE1 gene from the white yeast: The above sequence of amino acids is not by itself very informative. It is more important to compare the amino acid sequence of this protein in the red yeast to the amino acid sequence in the standard white yeast. This can be done using a program that was used before, BLAST. BLAST can be used to search databases for either DNA sequences (as in step 3) or protein sequences (what we are about to do). To do this, select “Copy Translation” under the “Edit” menu.

Now go back to the http : //www . y ea s tg e nom e . o r g web address and choose BLAST, or use your back button to back up one page. Delete any text that might be in the “Type or Paste Query Sequence” window. Next paste the amino acid sequence that you just copied into this window. You will now search the database for any protein from the standard (wild-type) white yeast whose amino acid sequence is identical or similar to the one that you just determined was encoded by the ADE1 gene of the red yeast. Under the heading of “Choose the appropriate BLAST program” select “BLASTP”. Under the next heading "Choose a Sequence Database,” select “Open Reading Frames (DNA or Protein)". These steps allow you to compare the red yeast p r o t e in sequence encoded by the ADE1 gene to all of the standard (wild-type) white yeast p r o t e ins in the database. Then select the "Run BLAST" option.

Results of the search: The results of this search are depicted in a way that is similar to step 3. The results are shown schematically below the query “number line” in the new window. The thicklines are color-coded to indicate the degree to which the query sequence and the target sequence match. In this case, there is only one “good” match shown in green—each blue line corresponds to an amino acid sequence whose similarity is so weak as to be of no or little significance. The green line runs the length of the “number line” indicating that the match extends for the entire length of

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the input sequence. The amino acid sequence of the Ade1 protein from the red yeast and the match to that sequence from the database are shown lower on the page---you can get there simply by clicking on the thick green line. At the top of this page is a title of the Subject sequence (the one that was found in the database). Notice that ADE1 is in the title (as well as a catalog number; this verifies that the sequence that has the best similarity to the Ade1 amino acid sequence from red yeast is the Ade1 protein from the white yeast). Just below the title of the Subject sequence you will see an alignment of letters, Scroll down until you see an alignment of letters, presented as sets with three lines of letters each. Again, each letter is the single letter abbreviation for an amino acid. Thus, there are 20 different letters that you could see. In this case, hash marks are not used to indicate amino acids that are identical between the two sequences. Instead the top line of each set of three is your query sequence – the protein encoded by the ADE1 gene from the red yeast. The bottom line of each set of three is the “Sbjct” sequence, the protein encoded by the white yeast DNA in the database. The middle line indicates the amino acids that are shared by both the query and Sbjct sequences – i.e., if both Sbjct and query have a “V” (for the amino acid valine) then a

“V” will be shown in the middle line. Gaps in the middle line indicate that the two sequences do not match at this position.

ii. Ar e t h e t w o ami n o a c id s e q u e n ce s id e n ti c al or a r e t h e r e di ff e r e n ce s ? I f th e r e a r e diff e r e n ce s th e n add r e s s the follo w ing qu e s tion s : w hat amino a c id numb e r in the c hain is diff e r e nt? P r o v ide the s ingle l e tt e r abb r ev iation f o r the name of the amino a c id that is p r e s e nt at that po s ition in the A D E 1 g e ne f r om the r e d ye a s t . The reading that y ou did la s t w ee k c at e go r i z e d mutations as "non s e n s e " , " mi ss e n s e ", " s il e nt", or "f r am e s hift". W hi c h c at e go r y do e s this fit into? L o o k ing at the table of s t r u c tu r e s of the amino a c ids p r o v id e d n e ar the b e ginning of this manual, d e s c r ibe the diff e r e n c e in the amino a c id that is alt e r e d — is it s imilar to, s lightly di f f e r e nt, or r adi c ally diff e r e nt in s t r u c tu r e f r om the o r iginal amino a c id? T hus is the mutation li ke ly to ha r m t he ability of the A d e 1 p r ot e in to c a rr y out its fun c tion? E x plai n . I s the diff e r e n c e in the amino a c id s e qu e n c e of the A d e 1 p r ot e in in the r e d s t r ain c ompa r e d to the w hite s t r ain a r e fl ec tion of a diff e r e n c e in the D N A s e qu e n c e b e t w ee n the A D E 1 g e ne of the r e d s t r ain c ompa r e d to the w hite s t r ain? E x plain y our an sw e r .

I f the t w o amino a c id s e qu e n ce s a r e id e nti c al, is it li ke ly that the diff e r e n c e in c olor b e t w ee n the r e d and w hite s t r ains is att r ibutable to a diff e r e n c e in the A D E 1 g e ne or c o rr e s ponding p r ot e in? E x plain.

6) Examining the database for a human gene that encodes a (weakly) similar protein that carries out the same function. We are using a lowly single-celled organism to learn about adenine biosynthesis---that is, how the critical compound adenine is made inside a cell. One reason to use such an organism is that it offers many advantages over more complex ones. For instance, it grows exceedingly fast. In 24 hours one cell will grow to more than 60,000 cells. This makes it simple to gather sufficient material on which to do experiments. In cases where one is examining different conditions for growth or absence of growth the experiment can be done in one day. Another advantage is the ability to treat the organism in a variety of ways. Consider the treatments you have done to the yeast during the last several weeks. It is unlikely that you could (or should) treat a vertebrate organism in the same way.

But do yeast cells have anything in common with cells from more complex organisms? We can make one attempt at answering this question by asking whether cells from humans have a protein that is similar to the protein encoded by the ADE1 gene of yeast. If the answer to this question is “yes” then it seems reasonable to conclude that both yeast cells and human cells make their adenine in similar ways. Furthermore, this would then suggest that studying adenine biosynthesis in yeast is an adequate way to begin to understand adenine biosynthesis in humans.

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In order to determine if human cells have a protein that is similar to the protein encoded by the ADE1 gene of yeast we will search a database that has all of the amino acid sequences of proteins that have been determined by scientists thus far in any type of organism. In your web browser click on the link that says “SGD locus page,” just above the ADE1 alignment. This should take you to the page titled ADE1/YAR015W that you saw last week. On the left side of the page find “Analyze Sequences” and select “BLASTP at NCBI”. A new page will load. This page will allow you to search a species-particular database for an amino acid sequence that is similar to the ADE1 protein of yeast. Like the earlier search of the yeast database, this site is using the program BLAST to find sequences similar to the Ade1 protein. Where the resulting page says “Choose search set” type “Homo sapiens” into the “Organism” window and then select the pulldown “Homo sapiens taxid:9606). Then click on the “BLAST” button. The page does a little rearranging at this point and it may take about 30 seconds for the search results to appear. The page may refresh one or more times. On the final results page most of the page background will be light blue.

Scroll down on the resulting page until you see the alignment of the yeast and the human protein. The top line of each set should be the yeast Ade1 protein sequence. The third line is the sequence of the human protein. The middle line has many blank spaces, these represent a place at which neither the human and yeast protein share similarity. Places in the middle line that have a character represent places in which there is identity between the human and yeast protein. A “+” character indicates a position in which the human and yeast protein have a similar but not identical amino acid. So the areas of the middle line with the highest density of characters are the areas that are the most similar between the yeast and human proteins. Note that the degree of similarity is not the same throughout the protein – the beginning (N- terminus), for example, is fairly similar between the two sequences (GK ++++YE+). The most similar regions are usually especially important in the function of the protein.

Examine the comparison. Near the end of the yeast protein amino acid sequence of ADE1 there is a sequence that is DKQFLRD. There is a similar sequence in the human proteinDKQSYRD. One could argue that the similarity between the yeast and human proteins is due to chance. But notice the “expected value” associated with the match between the yeast andhuman amino acid sequence. This value estimates the probability that such an amino acid sequence match could be found by random chance. This value is fairly low (4e-08; = one in 25 million) suggesting that there is some significance to this match.

Click on the link to the upper left of the alignment where it is written “G E N E I D: 10606 P A I C S .” This will take you to a page describing the human protein sequence. Notice the description of this protein includes “phosphoribosylaminoimidazole succinocarboxamide synthetase.” (Also there is a summary paragraph that says that this is involved in purine biosynthesis). This is an alternative name for the enzyme SAICAR synthetase---recall from last week that the name of the enzyme encoded by the yeast ADE1 gene is known as SAICAR synthetase but that it was also known as phosphoribosylaminoimidazole succinocarboxamide synthase. Thus we have found that the protein from humans that is similar in acid s e qu e n c e to the yeast Ade1 protein also has the same f un c tion – synthesizing the intermediate SAICAR in the pathway that manufactures adenine. Thus experiments investigating the yeast enzyme might tell us important things about the human enzyme.

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You have now completed an analysis of the ADE1 gene from the red yeast and compared it to the ADE1 gene from the white yeast. You have identified the base sequence of the ADE1 gene from the red yeast, you have identified the ADE1 base sequence from the standard (wild-type) white yeast, and you have compared these two base sequences. You have also determined the sequence of amino acids that is encoded by the ADE1 gene from the red yeast and compared it to theamino acid sequence of the protein encoded by the ADE1 gene from the white yeast. Finally, you have seen that a protein similar to the Ade1 protein, SAICAR synthetase, also can be found in humans. See the section at the end of the lab about what to discuss with your partner before youleave the lab.

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Below are the objectives and procedures for studying the ADE2 gene.

1) Examining the primary data of the ADE2 gene2) Converting the ADE1 sequence to that on the complementary DNA strand.3) Comparing the nucleotide sequence of the ADE2 gene from the red yeast to the ADE2 gene

from the white yeast4) Determining the amino acid sequence of the protein encoded by the ADE2 gene5) Comparing this amino acid sequence to the sequence from the standard (wild-type)

white strain6) Examining the database for a human gene that encodes a (weakly) similar protein that carries

out the same function

1) Examining the primary data of the ADE2 geneOpen up the file labeled “ADE2mr.ab1” using the program “4Peaks”. To do this, double-click onthe “4Peaks” icon, and under the pulldown heading of “File” select “Open”. Then select the file name “ADE2mr.ab1”. This computer file contains the sequence data we obtained from the PCR product generated with the ADE2 primer set in the previous lab session. This is the sequence of bases corresponding to the ADE2 gene from the red yeast strain. At this point you should see peaks of four different colors. There is also an X-axis that consists of a time unit of measurement, a nucleotide number, and a nucleotide (designated A, C, G, or T). There are a few things to note:

i) at many points there is only one peak and that the color of the peak determines the nucleotide indicated on the X-axis.

ii) at some points, there is more than one peak (this sequence actually has very few cases where this is true). This is due to technical problems and makes it difficult or impossible to predict the base that is at that position on the DNA strand. This is why there is occasionally a “N” instead of an A, C, G, or T. “N” designates that the identity of the base at this position was not unambiguously determined.

iii) it is technically more difficult to determine the base sequence near the beginning. In this case, the sequence is unreliable for the first few bases (for example, after the A base at position number 5 there should be a C, but the blue C peak is obscured by the larger green A peak). It is also technically difficult to determine the base sequence much more than about500 nucleotides from the start. Therefore we will limit our analysis to the base sequence between position 23 and 578.

Highlight (select) the characters between position 23 and 578 (from GTCCA to GGGTA) and then select "Crop" from the drop-down menu in the lower left hand corner of the window (the ‘gear’ icon). The sequence that remains in the 4Peaks window is now just the DNA between bases 23 and 578. Use “Save as” to rename this cropped file.

2) Converting the ADE2 sequence to that on the complementary DNA strand. Remember that DNA has two strands of nucleotides that run in opposite directions and are complementary. The sequence you have shows only one strand of the DNA, but because of the strict base pairing rules

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you can derive what the other strand must look like. For example, if the sequence of one strand is5’ CCACGT 3’ then the sequence of the complementary strand must be 5’ ACGTGG 3’. As luck would have it, the sequence you have received is actually the sequence of the gene on the template strand, read from the end of the gene to the beginning. In other words, the sequence is in reverse and complementary to the mRNA sequence that would be used to make a protein. In order to make the analysis of the ADE2 gene easier, we need to turn the sequence around so that it reads like the other DNA strand (called the “non-template strand”). We will use 4Peaks to make this change. We could do this manually, but it would be very tedious to write out the complementof hundreds of bases.

To convert the base sequence into the complementary base sequence, select “Flip” from the drop-down menu in the lower left hand corner of the window (the ‘gear’ icon). Note that the new sequence that is generated is simply the sequence of the second strand of the original DNA displayed 5’-> 3’.

3) Comparing the sequence of the ADE2 gene from the red yeast to the sequence of the ADE2 gene from the white yeast. In the following steps you will search SGD, the database that has the DNA sequence of all genes from the standard (wild-type) white yeast that you used last week, tosee if any have the same, or similar, sequence of bases as the sequence you have determined for theADE2 gene from the red yeast. To prepare for the comparison, copy the sequence by selecting “Copy Sequence” under the “Edit” menu.

a) Accessing the database via the internetOpen a web browser and type in the following address http://www . y ea s tg e nom e . o r g At this website, the entire base sequence of the standard (wild-type) white strain ofSaccharomyces cerevisiae is stored. This represents about 10 million base pairs. Nested somewhere in this 10 million base pairs is the sequence for the ADE2 gene (only about 2000 base pairs long). At the top of this web page within the purple bar, scroll your mouse over “Analyze” and select “B L A S T ”. This program will search the entire sequence of bases of the Saccharomyces cerevisiae in attempt to find some sequence that is either identical or similar to the “query” sequence (BLAST is an acronym for Basic L ocal Alignment S equencing Tool). The query sequence is that which you type or paste into the window, as in the following step. In this case, the query sequence is the ADE2 base sequence from the red yeast strain.

Under the heading “Choose the Appropriate BLAST Program” make sure that BLASTn is chosen. This is the DNA/RNA database. N stands for nucleotide).

Under the heading “Choose One or More Sequence Datasets” you should select “Open Reading Frames (DNA or Protein)”.

b) Searching the DNA database: Now place the cursor into the window above which is labeled“Type or Paste a Query Sequence” and paste the sequence that you copied in step 1. Use

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either the Paste command from the Edit menu or hold down the apple and “V” keys to paste. Next select the button “Run WU-BLAST”. The query sequence (the ADE2 sequence from the red yeast strain) is now being compared to all 10 million base pairs in the database of the standard (wild-type) white yeast strain to find the sequence or sequences thatare most similar.

c) Matches to the query DNA sequence: After a short time you will see the results of this search although it will be difficult to interpret. Near the top of the page you will see what looks like a number line. This line represents the sequence you put in, with the numbers indicating the nucleotide number. Under this numbered line you will see one or more colored thick lines (actually they are arrows). Each thick line represents a match to the query sequence. In this case there is one match (there is one line), represented by a thick purple arrow. If you mouse over this line you will see a label "YOR128C ADE2 SGDID:S000005654…". These characters are simply a database identifier tag, like a catalog number for an item in a catalog. Note that among the tags is the name “ADE2”. As the legend in the box indicates, the line is color-coded to signify how closely your query sequence matches the sequence found in the database. In addition, there is a p value in the label you just read (p=5.8e-122). This p value is ameasurement of how closely the two DNA sequences match; the higher the negative exponent the better the match. This is a very close match. In order to see how closely the two base sequences match scroll down; you should see lines of base sequence (strings of A, C, G, and T) and hash-marks between them. Note the similarity in the DNA sequence between the query (the ADE2 gene of the red yeast) and the DNA sequence found in the database (labeled "Sbjct"; this is the ADE2 gene of the white [wild-type] yeast). The query sequence starts with base number (coordinate) 1 and goes to coordinate 556; the "Sbjct" sequence starts at coordinate 280 andgoes to coordinate 835. Note that each line of sequence alternates between the Query and the Sbjct sequence. A hash-mark between the Query and Sbjct sequences indicates that the base sequence is identical between the Query and the Sbjct at that coordinate. The top portion of the screen will look like this:

d) But is it an exact match? Recall that our hypothesis is that the difference between the white and red yeast strain is attributable to difference in the base sequence of the ADE1 or ADE2 genes. Such a difference could be fairly small so check carefully to see if the Query sequence and the Sbjct sequence are exact matches.

i. Do y ou s e e any d i ff e r e n ce s ? A ny diff e r e n c e w ould m e an that the ba s e s e qu e n c e of the A D E 2 g e ne of the r e d ye a s t is diff e r e nt f r om the ba s e s e qu e n c e of the A D E 2 g e ne o f the w hite ye a s t. I f a diff e r e n c e w as ob s e r ve d, w hat t r e atm e nt of the ye a s t in w ee k 1 l e d to this di f f e r e n c e in the ba s e s e qu e n ce ? As a reminder, the base sequence in the database (the “Sbjct”) came from the standard (wild-type) white yeast.

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Su mm a r y of S t e p s 1 - 3. At this point you have determined the sequence of bases that make up the ADE2 gene from the red yeast, searched a database and compared the sequence of bases of the ADE2 gene from the white yeast to the ADE2 gene of the red yeast. In the next step you will use the base sequence of the ADE2 gene from red yeast to determine the amino acid sequence of the protein encoded by the ADE2 gene from the red yeast. You will then compare the amino acid sequences from the red and white yeasts.

4) Determining the amino acid sequence of the protein encoded by the ADE2 gene from the red yeast. In this section you will use the program 4Peaks to determine the amino acid sequence of the protein encoded by the ADE2 gene.

a) What’s in a DNA sequence? Recall our definition of a gene: a gene is a segment of DNA that codes for a protein. As was the case last week, the computer shows you the sequence of the non-template strand of DNA, the strand that looks like the mRNA, except for the presence of T’s instead of U’s. This strand can be directly analyzed using a genetic code table to decipher what amino acids are encoded. The DNA code is written in groups of three bases. Note that the “reading frame” of the sentence is quite important. For instance, if I wrote a sentence of three letter words in the midst of a random assortment of letters it would be nonsense unless you started in the correct reading frame. The three reading frames of the sentence TETHECATATETHEBIGRATTT would be:

Reading frame (1) TET HEC ATA TET HEB IGR ATT T(groups of three start at the first letter)

Reading frame (2) T ETH ECA TAT ETH EBI GRA TTT(groups of three start at the second letter)

Reading frame (3) TE THE CAT ATE THE BIG RAT TT(groups of three start at the third letter)

There is only one of three reading frames in which you will be able to decipher the meaning of this sentence. So in this example, unless you start reading in the “3rd” reading frame, this string of letters would have no meaning. Likewise, to decipher the amino acid sequence of a gene the information must be read in the correct “reading frame”. The computer program 4Peaks will examine each of the three possible reading frames to determine which one has a “sentence”. In this case a sentence means a long string of amino acids that are encoded without any “stop” codes. Recall from the homework assignment that you did last week that some 3- letter combinations represent a “stop”---that is, rather than an amino acid this stop signal results in the termination of translation. The codes for stop are TGA, TGG, and TAG.

b) What and where is an ORF? An “ORF” is an abbreviation for an open r eading frame and is metaphorically the same as finding a sentence in the midst of a number of letters strung together. The 4Peaks program has tools designed for this purpose. Return to 4Peaks. Click on the Show Translation icon in the lower right hand corner of the window. This operation will “decode” (translate) your DNA sequence into an amino acid sequence by analyzing the nucleotide sequence in groups of three, noting any cases of stop codes as asterisks (*). Click on the “Set Translation Frame” icon in the lower right hand corner,

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to select a different reading frame. An open reading frame (orf) will contain no asterisks. This means that this segment of DNA has the potential to code for a protein because there are no stops, just amino acids, encoded.

c) The amino acid sequence of the protein encoded by the ADE2 gene of the red yeast. If you do see an ORF this indicates the potential for this sequence of bases to encode a protein. This is the sequence of amino acids in the protein encoded by the red yeast ADE1 DNA that you sequenced. The amino acids are given in one-letter abbreviations. For example, the amino acid serine is abbreviated with the letter S and the amino acid leucine is abbreviated with the letter L. If you have succeeded then the last five amino acids using the one letter abbreviation should be "PHNSG".

5) Comparing the amino acid sequence encoded by the ADE2 gene from the red yeast to the amino acid sequence of the protein encoded by the ADE2 gene from the white yeast: The above sequence of amino acids is not by itself very informative. It is more important to compare the amino acid sequence of this protein in the red yeast to the amino acid sequence in the standard white yeast. This can be done using a program that was used before, BLAST. BLAST can be used to search databases for either DNA sequences (as in step 4) or protein sequences (what we are about to do). To do this, select “Copy Translation” under the “Edit” menu.

Now go back to the http : //www . y ea s tg e nom e . o r g web address and choose BLAST, or use your back button to back up one page. Delete any text that might be in the “Type or Paste Query Sequence” window. Next paste the amino acid sequence that you just copied into this window. You will now search the database for any protein from the standard (wild-type) white yeast whose amino acid sequence is identical or similar to the one that you just determined was encoded by the ADE2 gene of the red yeast. Under the heading of “Choose the appropriate BLAST program” select “BLASTP”. Under the next heading "Choose a Sequence Database,” select “Open Reading Frames (DNA or Protein)". These steps allow you to compare the amino acid sequence of the protein encoded by the ADE2 gene of the red yeast to all of the standard (wild-type) white yeast p r ot e ins in the database. Then select the "Run WU-BLAST" option.

Results of the search: The results of this search are depicted in a way that is similar to step 3. The results are shown schematically below the query “number line” in the new window. The thicklines are color-coded to indicate the degree to which the query sequence and the target sequence match. In this case, there is only one “good” match shown in green (labeled YOR128C)—each blue line corresponds to an amino acid sequence whose similarity is so weak as to be of no or little significance. The amino acid sequence of the Ade2 protein from the red yeast and the match to that sequence from the database are shown lower on the page---you can get there simply by clicking onthe thick green line. At the top of this page is a title of the Subject sequence (the one that was foundin the database). Notice that ADE2 is in the title (as well as a catalog number; this verifies that the sequence that has the best similarity to the Ade2 amino acid sequence from red yeast is the Ade2 protein from the white yeast). Just below the title of the Subject sequence you will see an alignment of letters presented as sets with three lines of letters each. Again, each letter is the single letter abbreviation for an amino acid. Thus, there are 20 different letters that you could see. In this case, hash marks are not used to indicate amino acids that are identical between the two sequences. Instead the top line of each set of three is your query sequence – the protein encoded by the ADE2 gene from the red yeast. The bottom line of each set of three is the “Sbjct” sequence, the protein encoded by the white yeast DNA in the database. The middle line indicates the amino acids thatare shared by both the query and Sbjct sequences – i.e., if both Sbjct and query have a “V” (for the

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amino acid valine) then a “V” will be shown in the middle line. Gaps in the middle line indicate that the two sequences do not match at this position.

ii. Ar e t h e t w o ami n o a c id s e q u e n ce s id e n ti c al or a r e t h e r e di ff e r e n ce s ? I f th e r e a r e diff e r e n ce s th e n add r e s s the follo w ing qu e s tion s : w hat amino a c id numb e r in the c hain is diff e r e nt? P r o v ide the s ingle l e tt e r abb r ev iation f o r the name of the amino a c id that is p r e s e nt at that po s ition in the A D E 2 g e ne f r om the r e d ye a s t. T he reading that y ou did la s t w ee k c at e go r i z e d mutations as "non s e n s e " , " mi ss e n s e ", " s il e nt", or "f r am e s hift". W hi c h c at e go r y do e s this fit into? L o o k ing at the table of s t r u c tu r e s of the amino a c ids p r o v id e d n e ar the b e ginning of this manual, d e s c r ibe the diff e r e n c e in the amino a c id that is alt e r e d — is it s imilar to, s lightly di f f e r e nt, or r adi c ally diff e r e nt in s t r u c tu r e f r om the o r iginal amino a c id? T hus is the mutation li ke ly to ha r m t he ability of the A d e 2 p r ot e in to c a rr y out its fun c tion? E x plai n . I s the diff e r e n c e in the amino a c id s e qu e n c e of the A d e 2 p r ot e in in the r e d s t r ain c ompa r e d to the w hite s t r ain a r e fl ec tion of a diff e r e n c e in the D N A s e qu e n c e b e t w ee n the A D E 2 g e ne of the r e d s t r ain c ompa r e d to the w hite s t r ain? E x plain y our an sw e r .

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I f the t w o amino a c id s e qu e n ce s a r e id e nti c al, is it li ke ly that the diff e r e n c e in c olor b e t w ee n the r e d and w hite s t r ains is att r ibutable to a diff e r e n c e in the A D E 2 g e ne or c o rr e s ponding p r ot e in? E x plain.

6) Examining the database for a human gene that encodes a (weakly) similar protein that carries out the same function. We are using a lowly single-celled organism to learn about adenine biosynthesis---that is, how the critical compound adenine is made inside a cell. One reason to use such an organism is that it offers many advantages over more complex ones. For instance, it grows exceedingly fast. In 24 hours one cell will grow to more than 60,000 cells. This makes it simple to gather sufficient material on which to do experiments. In cases where one is examining different conditions for growth or absence of growth the experiment can be done in one day. Another advantage is the ability to treat the organism in a variety of ways. Consider the treatments you have done to the yeast during the last several weeks. It is unlikely that you could (or should) treat a vertebrate organism in the same way.

But do yeast cells have anything in common with cells from more complex organisms? We can make one attempt at answering this question by asking whether cells from humans have a protein that is similar to the protein encoded by the ADE2 gene of yeast. If the answer to this question is “yes” then it seems reasonable to conclude that both yeast cells and human cells make their adenine in similar ways. Furthermore, this would then suggest that studying adenine biosynthesis in yeast is an adequate way to begin to understand adenine biosynthesis inhumans.

In order to determine if human cells have a protein that is similar to the protein encoded by the ADE2 gene of yeast we will search a database that has all of the amino acid sequences of proteins that have been determined by scientists thus far in any type of organism. In your web browser click on the link that says “SGD locus page,” just above the ADE2 alignment. This should take you to the page titled ADE2/YOR128C that you saw last week. On the left side of the page find “Analyze Sequence.” Select “BLASTP at NCBI”. The NCBI database holds sequences from all types of organisms. Like the earlier search of the yeast database, this site will use the program BLAST to find sequences similar to the Ade2 protein. To limit our search for similar proteins to humans, under “Organism” in the “Choose Search Set” section type “Homo sapiens” and then select the pulldown “Homo sapiens (taxid:9606).” Now go to the line “Filters and Masking” and click on the box that has a check mark to deselect this option. Then hit the blue “BLAST” button at the bottom of the page. The search is now on for any amino acid sequence from Homo sapiens that bears a resemblance to the amino acid sequence encoded by ADE2. The page will refresh one or more times. The final page will have a background that is mostly light blue.

When the search is finished a new page will appear. Scroll down the page until the heading“Alignments.” If all went well you should see the top match with the identifier “Gene ID10606PAICS”. Click on this link and you will learn about the human gene that has the highest similarity to the yeast ADE2 gene. The title of this page includes the text “phosphoribosylaminoimidazole carboxylase.” Recall from last week that the name of the

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enzyme encoded by the yeast ADE2 gene was phosphoribosylaminoimidazole carboxylase or AIR carboxylase. Thus we have found that the protein from humans that is most similar in acid s e qu e n c e to the yeast Ade2 protein also has the same f un c tion – adding a carboxyl group onto an intermediate (phosphoribosylaminoimidazole) in the pathway that manufactures adenine. Also notice under the Summary comments that this human enzyme seems to also participate in purine biosynthesis. Thus experiments investigating the yeast enzyme might tell us important things about the human enzyme.

Hit the Back button on your browser to return to the alignment page. Scroll down on the resulting page until you see the alignment of the yeast and the human protein. The top line of each set should be the query sequence – the yeast Ade2 protein sequence. The third line of each set is the sequence of the human protein. The middle line contains some letters and some plus signs. Note that the letters appear at each position where the two sequences are identical. Although it is less obvious unless you know the structures of the amino acids, a “+” appears where the amino acids are similar to each other in structure – an L and a V, for example. Sothe areas with the highest density of letters and + symbols are the areas that are the most similarbetween the yeast and human proteins.

Examine the comparison. One could argue that the similarity between the yeast and human proteins is due to chance. But notice the “expect value” associated with the match between the yeast and human amino acid sequence. This value estimates the probability that such an amino acid sequence match could be found by random chance. This value (1e-04) is fairly low suggesting that there might be some significance to this match. Also note that the degree of similarity is not the same throughout the protein – the region ++MGS-SDL near the beginning of the alignment, for example, is quite similar between the two sequences. The most similar regions are usually especially important in the function of the protein.

Before you leave the lab, exchange data with your lab partner by filling out the sheet on the next page and having the instructor check it. Having data for both the ADE1 and ADE2 genes will be crucial in order to write the lab report.

SUMMARY: You and your partner have examined the base sequence of two genes ADE1 and ADE2 from the red yeast. The reason that you examined these two genes is that previous research has indicated that changes in either of these two genes can give rise to yeast cells that have a red pigment. Could this be true in the case of the red yeast that you observed after ultraviolet light treatment? You have compared the sequence of the ADE1 gene and the ADE2 gene from the red yeast to the corresponding genes from the white yeast. Were there differences between the red and white yeast in either ADE1 or ADE2? In b o t h ADE1 and ADE2? You have also examined the sequence of amino acids encoded by the ADE1 and ADE2 genes from the red yeast and compared these sequences to the corresponding amino acid sequences from the white yeast. Were there amino acid sequence differences between proteins encoded by the ADE1 a n d ADE2 genes in red and white yeast?

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QUESTIONS FOR DISCUSSION: B e f o r e you l e ave t h e la b , please meet with your lab partner to answer the following questions. Write the answers below and keep this page for your future reference after showing it to the instructor or TA.

1) You should have seen some difference between the red yeast DNA and the white yeast DNA in the database. What difference did you see? Write here:

The gene that had the difference (was it ADE1 or ADE2?):

The position of the base that was changed and what that change was: Base number:

Base seen at that position in white yeast DNA in the database.

Base seen at that position in red yeast DNA that you sequenced:

2) Was there a corresponding alteration in the protein encoded by the altered gene? What difference did you see? Write here:

The position of the amino acid that was altered and the type of alteration: Amino acid number:

Amino acid seen at that position in white yeast in the database:

Amino acid seen at that position in the red yeast:

Nature of the mutation (silent, missense, nonsense, frameshift):

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